"Vanishing inner product implies orthogonality" Is it a definition or theorem?

Question

The definition of inner product is $\langle u,v\rangle=u_1\bar{v_1}+\cdots+u_n\bar{v_n}$. Two vectors $u,v \in V$ are said to be orthogonal if $\langle u,v\rangle=0$, where $V$ is complex vector space. But why is it true? A lot of books just present this fact without giving any proof. Why such an expression $u_1\bar{v_1}+\cdots+u_n\bar{v_n}=0$ will lead to the fact that $u$ and $v$ are orthogonal? Is it an intuition behind the expression $u_1\bar{v_1}+\cdots+u_n\bar{v_n}$ that can explain orthogonality?

I read some proof about it but they don't seem correct. For example, some said that $\langle u,v\rangle=|u||v|\cos\theta$ and thus $\theta=90^\circ$. But why are they equivalent?

Some use the Pythagoras theorem, $$\begin{array}{rl} |u+v|^2 &=\langle u+v,u+v\rangle\\ &=|u|^2+2\langle u,v\rangle+|v|^2 \end{array}$$ So dot product $= 0$ implies orthogonality, but this only works for real vector space. So why exactly inner product $= 0$ implies orthogonality?

Answer 1

Just to clarify the excellent other answers and Willie Wong's illucid comment:

Orthongonal := inner product equals zero--- a linear algebra term about vectors (applies to all vector spaces); is a definition.

Perpendicular := intersect at a right angle--- a geometry term about lines, planes or higher dimensional hyper-planes (applies to Euclidean planes and spaces); is a definition.

$\mathbb R^n$ representing Euclidean $n$-space and vectors in $\mathbb R^n$ representing Euclidean lines or planes or hyper planes; is an interpretation.

In $\mathbb R^n$, orthogonal if and only if perpendicular; is a theorem (which is provable by the Pythagorean Theorem)

(-- and which is frequently not considered an important theorem and ignored, as many [most?] texts do not considered classic geometry to be important; instead all geometry is only interpreted in linear algebra terms, in which "perpendicular" is not used.)

(In a way, asking why orthogonal/perpendicular means inner product is $0$, is like asking why $\{(t, t(x)+a|x \in \mathbb R^n\}$ is a line. It's a result of interpreting classical geometry into analytical terms, and then deciding the classical interpretation is no longer pertinent and, from then on, only using the analytic interpretation as the very basis and definition of geometry instead.)

Answer 2

The geometrical interpretation of the inner product have made usage of some of the terminology of geometry in vector spaces. I guess that the concept of orthogonality is one of them, and as the dot product defines orthogonality when it is equal to zero, then orthogonality in a vector space is defined as zero inner product.

Read what this wikipedia article says about it: https://en.wikipedia.org/wiki/Inner_product_space (in orthogonality section)

Answer 3

As Willie Wong mentioned in his comment, the definition we take in general vector spaces is that two elements of the vector space are orthogonal if their inner product is zero. However, in $\Bbb R^2$, we can see motivation for this definition in terms of the Euclidean dot product. Suppose $a, b\in \Bbb R^2$ and $a\cdot b = 0$. Then, $$ a_1b_1 + a_2b_2 = 0 \iff a_1/b_2 = -a_2/b_1. $$ This is a manifestation of the fact that if two lines have slopes that are negative reciprocals of one another, the two lines are perpendicular in $\Bbb R^2$. Hence, two lines are perpendicular in the sense of Euclidean geometry provided that their dot product is zero. This is motivation for the general definition of orthogonality in more abstract vector spaces.